Question

1. k20q1

nasa operates a drop tower in which they test the response of materials to situations in which the only force which acts upon them is gravity. objects are dropped from rest and free fall for 5.12 seconds through the drop tower.
a. determine the distance which the objects free fall through the tower.
b. to gain a feel for the height of the drop tower, convert this falling distance to units of feet. (given: 1 meter = 3.28 feet)

Explanation:

Step1: Use the free - fall distance formula

The formula for the distance \(s\) of an object in free - fall (starting from rest, \(u = 0\)) is \(s=ut+\frac{1}{2}gt^{2}\). Since \(u = 0\), the formula simplifies to \(s=\frac{1}{2}gt^{2}\). The acceleration due to gravity \(g = 9.8\ m/s^{2}\), and \(t = 5.12\ s\).

$$s=\frac{1}{2}\times9.8\times(5.12)^{2}$$

Step2: Calculate the value of \(s\)

First, calculate \((5.12)^{2}=5.12\times5.12 = 26.2144\). Then \(\frac{1}{2}\times9.8\times26.2144=4.9\times26.2144 = 128.45056\approx128.5\ m\)

Step3: Convert meters to feet

Given the conversion factor \(1\ m=3.28\ ft\). If \(s = 128.5\ m\), then to convert to feet, we use the formula \(d_{ft}=s\times3.28\)

$$d_{ft}=128.5\times3.28=(128 + 0.5)\times3.28=128\times3.28+0.5\times3.28$$

$$128\times3.28=(130 - 2)\times3.28=130\times3.28-2\times3.28=426.4-6.56 = 419.84$$

$$0.5\times3.28 = 1.64$$

$$d_{ft}=419.84 + 1.64=421.48\approx421.5\ ft$$

Answer:

a. \(128.5\ m\)
b. \(421.5\ ft\)